A piece of wood from a recently cut tree shows \(20\,decays\) per minute. A wooden piece of same size placed in a museum ( obtained from a tree cut many years back) shows \(2\,decays\) per minute. If half life of \(C^{14}\) is \(5730\, years\), then age of the wooden piece placed in the museum is approximately ........... \(years\)

\(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=2\) decays/min

\(\mathrm{T}_{1 / 2}=5730 \mathrm{years}\)

As we know,

\(\mathrm{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t}\)

\(\log \frac{N_{0}}{N}=\lambda t\)

\(\therefore \mathrm{t}=\frac{1}{\lambda} \log \frac{\mathrm{N}_{0}}{\mathrm{N}}\)

\( = \frac{{2.303 \times {{\text{T}}_{1/2}}}}{{0.693}} \times {\text{Lo}}{{\text{g}}_{10}}\frac{{{{\text{N}}_0}}}{{\text{N}}}\)

But \(\frac{\frac{d N_{0}}{d t}}{\frac{d N}{d t}}=\frac{N_{0}}{N}=\frac{20}{2}=10\)

\(\therefore \,\,t = \frac{{2.303 \times 5730}}{{0.693}} \times 1\)

\(=19039\) years